So Borel Sigma Field.

So this is the minimal Sigma Field which contains all open intervals.

So what kind of subsets are in?

The bottle Sigma field so notation is.

Beer.

Bordel

Sigma Field.

So first of all, if it contains open intervals, right?

And finite collections. Infinite collections of.

Open intervals.

Well, let's say denumerable.

Open intervals.

Right because Sigma Field?

Sigma field must contain all.

Countable or denumerable?

Collections of of its elements, hotels.

Intervals closed interval AB. Any closed interval belongs to B as well.

Why?

Becausw

Becausw Interval.

Which is closed.

Is intersection of the open intervals.

A minus Epsilon.

Be.

Plus insurance if we take intersection.

Overall.

Absolute or the shape? 1 / N then it will be exactly closed interval is this clear?

For my understanding, the boreal field contains all of the possibility in this interval, is it contains every number, every single number, and every interval? Is it every single number? Every single number means a point.

Particular boiling point belongs to B.

Every countable.

Collection of points.

A1A2

search countable set belongs to B.

Because it is countable union.

Closed intervals.

Belong to be OK. Thank you. It's clear all combinations, open intervals, closed intervals, particular points and so on.

So what is?

More challenging

the question.

Alright, more challenging question.

Do they exist?

Fits

X from.

From

radio numbers.

Search that.

X is not in borrow Sigma Field.

Well that is.

Very difficult question.

Well, the answer is yes.

But such sets are very complicated.

I'm not going to construct them. Maybe later I will tell you something about that, but in the textbooks, usually in appendices, as a difficult material, such sets are described.

So the boreal Sigma Field is very rich.

It contains very many different subsets.

Like here, singletons, open intervals, closed intervals. Are there unions?

Will denumerable compute combinations, but not says from real numbers are borrow?

Unless be is very rich.

So I think that is.

Enough.

For this.